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The percent dissociation of a base $\ce{X}$ in a $0.002\ \mathrm{M}$ solution is $4.7 \cdot 10^{-3}\ \%$. What is the pH of the solution and what is the $\mathrm{p}K_\text{b}$ of base $\ce{X}$? The autoionization of water may not be neglected.

Attempt at solution: We have the following reactions: $$ \ce{X + H2O(l) <=> XH+ + OH-} \\ \ce{H2O(l) <=> H+ + OH-}$$

Since the percent dissociation is $4.7 \times 10^{-3}\ \%$, this enables us to find the concentration dissociated: $$ \frac{x}{0.002\ \mathrm{M}} \cdot 100 = 4.7 \times 10^{-3},$$ so $x = 9.4 \times 10^{-8}$. Now I did: $$ K_\text{a} = \frac{x^2}{0.002 - x}$$ and I let $(0.002 - x) \approx 0.002$ since $x$ is so small. But this gives me the wrong answer, and it seems I haven't taken the autoionization of water into account yet. How should I solve this problem then?

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In such an solution, since the autoionization of water may not be neglected, you should expect $x=9.4 \times 10^{-8}$ do not mean $[\ce{OH-}]=9.4 \times 10^{-8}$. So you can only leave it unknown.

Then, $K_\text{b} = \frac {[\ce{XH+}][\ce{OH-}]}{[\ce{X}]} \approx \frac {9.4 \times 10^{-8} [\ce{OH-}]}{0.002} = 4.7 \times 10^{-5}[\ce{OH-}]$.

However, we still do not know what $[\ce{OH-}]$ is.

If only to consider the autoionization of water balanced, you will get

$[\ce{OH-}][\ce{H+}] = K_\text{w} = 1 \times 10^{-14}$

and this is still not enough.

The answer can only consider by the fact that the solution will be electrically neutral,

$[\ce{OH-}]=[\ce{H+}] + [\ce{XH+}] = \frac {1 \times 10^{-14}}{[\ce{OH-}]} + 9.4 \times 10^{-8}$

$[\ce{OH-}]^2 - 9.4 \times 10^{-8} [\ce{OH-}] - 1 \times 10^{-14}=0$

Since $\Delta = (9.4 \times 10^{-8})^2 + 4 \times 10^{-14} \approx 4 \times 10^{-14}$,

$[\ce{OH-}] \approx \frac {9.4 \times 10^{-8} + 2 \times 10^{-7}}{2} = 2.47 \times 10^{-7}$.

Thus, $\mathrm{pH} = -\log \left(\frac {1 \times 10^{-14}}{2.47 \times 10^{-7}}\right) = 7.39$

$\mathrm{p}K_\text{b} = -\log (4.7 \times 10^{-5} \times 2.47 \times 10^{-7}) = 10.94$.

Sometimes the “electrically neutral” trait will be very useful, this is a good example.

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  • $\begingroup$ pkb = 10.94 ? At pH = 7.39, this would imply that the concentration of the base is much higher than the concentration of the corresponding acid. However, it is stated in the question that the percent dissociation of the base was only 4.7*10^-3 %. These things do not match. $\endgroup$ – Bive Apr 3 '17 at 14:51
  • $\begingroup$ The method applied in order to solve this problem is in principle ok. However, the statements given are not correct, leading to wrong conclusions in the calculations. [XH+] is not equal to 9.4*10^-8 (M), but to 0.002 - 9.4*10^-8 = 0.01999... (M). [X] is not equal to 0.002 (M), but to 9.4*10^-8 (M). $\endgroup$ – Bive Apr 3 '17 at 21:18
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Let us call the base for A- and its corresponding acid for HA.

Ctot = [HA] + [A-] = 0.002 (M) => [HA] = 0.002 - [A-].

Equilibrium: A- + H2O = HA + OH-

[A-]/( [A-] + [HA]) = 4.7*10^-3/100 = 4.7*10^-5 =>

[A-] / ([A-] + 0.002 - [A-]) = 4.7*10^-5 =>

[A-] = 9.4*10^-8 (M) => [HA] = 0.002 - 9.4*10^-8 = 0.001999906 (M)

The charge balance of the system is: [H3O+] = [OH-] + [A-]

To compensate for the autoprotolysis of water we multiply all terms of the charge balance with [H3O+] and get:

[H3O+]^2 = [H3O+] * [OH-] + [H3O+] * [A-] =

10^-14.0 + [H3O+] * [A-]

=> [H3O+]^2 - 9.4*10^-8 *[H3O+] - 10^-14.0 = 0

[H3O+] = 1.574 * 10^-7 => pH = 6.802

At equilibrium:

pH = pka + log [A-]/[HA] =>

pka = pH - log [A-]/[HA] = 6.802 - log (9.4*10^-8/0.001999906) => pka = 6.802 + 4.327 = 11.129.

pka + pkb = 14.0 => pkb = 14.0 - 11.129 = 2.87

Answer:

pH ≈ 6.8

pkb ≈ 2.9

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